Consider the two sets: A = {m in R : text{bot | vedclass

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(A) For the roots of the quadratic equation $x^{2} - (m+1)x + m+4 = 0$ to be real,the discriminant $D$ must be greater than or equal to $0$.
$D = (m+

Vedclass · Accepted Answer

$A - B = (-\infty, -3) \cup [5, \infty)$

Vedclass · Answer

(A) For the roots of the quadratic equation $x^{2} - (m+1)x + m+4 = 0$ to be real,the discriminant $D$ must be greater than or equal to $0$.
$D = (m+1)^{2} - 4(m+4) \geq 0$
$m^{2} + 2m + 1 - 4m - 16 \geq 0$
$m^{2} - 2m - 15 \geq 0$
$(m-5)(m+3) \geq 0$
Thus,$m \in (-\infty, -3] \cup [5, \infty)$,so $A = (-\infty, -3] \cup [5, \infty)$.
Given $B = [-3, 5)$.
Now,evaluate the options:
$A - B = (-\infty, -3) \cup [5, \infty)$ (True)
$A \cap B = \{-3\}$ (True)
$B - A = (-3, 5)$ (True)
$A \cup B = (-\infty, -3] \cup [-3, 5) \cup [5, \infty) = (-\infty, \infty) = R$ (True)
Wait,checking the options again: $A - B = (-\infty, -3) \cup [5, \infty)$. Since $-3 \in B$,it is removed from $A$. So $A - B = (-\infty, -3) \cup [5, \infty)$. This is true.
All options provided are actually true. However,if we re-examine $A \cup B$,it is $R$. If we re-examine $B - A$,it is $(-3, 5)$.
Given the standard nature of this problem,there might be a typo in the question's options. Assuming the question asks for the false statement,and all are true,we identify that $A \cup B = R$ is true,$A \cap B = \{-3\}$ is true,$B - A = (-3, 5)$ is true,and $A - B = (-\infty, -3) \cup [5, \infty)$ is true. Since all are true,the question is flawed.

Consider the two sets: $A = \{m \in R : \text{both the roots of } x^{2} - (m+1)x + m+4 = 0 \text{ are real}\}$ and $B = [-3, 5)$. Which of the following is not true?

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